Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{k^2 - 16k + 60}{2k^2 - 16k - 40} \div \dfrac{2k^2 - 14k}{k^2 + 2k} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{k^2 - 16k + 60}{2k^2 - 16k - 40} \times \dfrac{k^2 + 2k}{2k^2 - 14k} $ First factor out any common factors. $q = \dfrac{k^2 - 16k + 60}{2(k^2 - 8k - 20)} \times \dfrac{k(k + 2)}{2k(k - 7)} $ Then factor the quadratic expressions. $q = \dfrac {(k - 10)(k - 6)} {2(k - 10)(k + 2)} \times \dfrac {k(k + 2)} {2k(k - 7)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac { (k - 10)(k - 6) \times k(k + 2)} { 2(k - 10)(k + 2) \times 2k(k - 7)} $ $q = \dfrac {k(k - 10)(k - 6)(k + 2)} {4k(k - 10)(k + 2)(k - 7)} $ Notice that $(k - 10)$ and $(k + 2)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {k\cancel{(k - 10)}(k - 6)(k + 2)} {4k\cancel{(k - 10)}(k + 2)(k - 7)} $ We are dividing by $k - 10$ , so $k - 10 \neq 0$ Therefore, $k \neq 10$ $q = \dfrac {k\cancel{(k - 10)}(k - 6)\cancel{(k + 2)}} {4k\cancel{(k - 10)}\cancel{(k + 2)}(k - 7)} $ We are dividing by $k + 2$ , so $k + 2 \neq 0$ Therefore, $k \neq -2$ $q = \dfrac {k(k - 6)} {4k(k - 7)} $ $ q = \dfrac{k - 6}{4(k - 7)}; k \neq 10; k \neq -2 $